Let X be a random sample representing the score of a student in the American Chemical Society Examination.
P(450 < x < 500) = P((450 - 500)/(90/sqrt(25)) < z < (500 - 500)/(90/sqrt(25))) = P(-2.778 < z < 0) = P(z < 0) - P(z < -2.778) = P(z < 0) - [1 - P(z < 2.778)] = P(z < 0) + P(z < 2.778) - 1 = 0.5 + 0.99727 - 1 = 0.49727 = 49.7%
To solve the problem we will find the number of videos bought by Jerome and the total cost of all the videos.
The system of equations that model this information are:
- 15 = x + y
- 164 = 18x + 9y.
<h2>Explanation</h2>
Given to us
- Total amount spent by Jerome = $164
- Total videos bought by Jerome = 15
- Number of classic videos bought = y
- Number of new release videos bought = x
- Cost of a classic video = $8
- Cost of a new release video = $19
<h3>Total cost of all new release videos</h3>
= Number of new releases bought x Cost of a new release video
![= x \times \$19\\ =19x](https://tex.z-dn.net/?f=%3D%20x%20%5Ctimes%20%5C%2419%5C%5C%0A%3D19x)
<h3>Total cost of all classic videos</h3>
= Number of classic videos bought x Cost of a classic video
![= y \times \$8\\ =8 y](https://tex.z-dn.net/?f=%3D%20y%20%5Ctimes%20%5C%248%5C%5C%0A%3D8%20y)
<h2>Total videos bought by Jerome</h2>
Total videos bought by Jerome
= Number of new releases bought + Number of classic videos bought
15 = x + y
<h2>Total amount spent by Jerome </h2>
Total amount spent by Jerome
= Total cost of all new release videos + Total cost of all classic videos
$164 = 19x + 8y
164 = 19x +8y
Hence, the system of equations that model this information are:
- 15 = x + y
- 164 = 18x + 9y.
Learn more about System of Equations:
brainly.com/question/12895249
<span>Write an equation in the form
y
=
m
x
+
b
y=mx+b for the following table:
x y
-10 -21
-8 -15
-6 -9
-4 -3
-2 3
0 9
2 15
4 21
</span>
Answer:
AB and CD would be 30
AC and BD would be 16
Step-by-step explanation:
The scale factor is what you multiply the sides by. Dilate means wider or larger so you multiply each side by the scale factor of 2 to get the higher scaled rectangle.